# Two different analytical tests can be used to determine the impurity level in steel alloys. Nine specimens are tested using both procedures, and the results are shown in the tabulation.Find the 95\% confidence interval for the mean difference of the tests, and explain it.

Two different analytical tests can be used to determine the impurity level in steel alloys. Nine specimens are tested using both procedures, and the results are shown in the following tabulation.
$\begin{array}{|ccc|}\hline Specimen& Test-1& Test-2\\ 1& 1.2& 1.4\\ 2& 1.6& 1.7\\ 3& 1.5& 1.5\\ 4& 1.4& 1.3\\ 5& 1.8& 2.0\\ 6& 1.8& 2.1\\ 7& 1.4& 1.7\\ 8& 1.5& 1.6\\ 9& 1.4& 1.5\\ \hline\end{array}$
Find the $95\mathrm{%}$ confidence interval for the mean difference of the tests, and explain it.

You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Cheyanne Leigh

Step 1
$\begin{array}{|cccccc|}\hline Test-1\left(X\right)& Test-2\left(Y\right)& \left(x-\overline{x}\right)& \left(x-\overline{x}{\right)}^{2}& \left(y-\overline{y}\right)& \left(y-bary{\right)}^{2}\\ 1.2& 1.4& -0.31111& 0.09679& -0.24444& 0.05975\\ 1.6& 1.7& 0.08889& 0.00790& 0.05556& 0.00309\\ 1.5& 1.5& -0.01111& 0.00012& -0.14444& 0.02086\\ 1.4& 1.3& -0.11111& 0.01235& -0.34444& 0.11864\\ 1.8& 2.0& 0.28889& 0.08346& 0.35556& 0.12642\\ 1.8& 2.1& 0.28889& 0.08346& 0.45556& 0.20753\\ 1.4& 1.7& -0.11111& 0.01235& 0.05556& 0.00309\\ 1.5& 1.6& -0.01111& 0.00012& -0.04444& 0.00198\\ 1.4& 1.5& -0.11111& 0.01235& -0.14444& 0.02086\\ \hline\end{array}$
Step 2
The mean is given by
$\stackrel{―}{x}=\frac{1}{n}\sum x$
Test 1 NKS $\stackrel{―}{x}=\frac{13.6}{9}=1.511$
Test 2
$\stackrel{―}{y}=\frac{14.8}{9}=1.644$
The standard deviation is given by
$\sigma =\sqrt{\frac{1}{n-1}\sum {\left(x-\stackrel{―}{x}\right)}^{2}}$
Test 1
$\sigma =\sqrt{\frac{0.30889}{8}}=0.1965$
Test 2
${s}_{2}=\sqrt{\frac{0.5622}{8}}=0.2651$
The $95\mathrm{%}$ CI is given by

$C.I.=\left(1.511-1.644\right)±2.306\sqrt{\frac{{\left(0.1965\right)}^{2}}{9}+\frac{{\left(0.2651\right)}^{2}}{9}}$

It shows that the true difference between the mean of test 1 and test 2 will lie within -0.387 and 0.120.